Decay Problem

**dwat** · June 7th 2009, 05:43 PM

"The half-life of radioactive carbon (14) is 5730 years. If the amount of radioactive carbon left over after 1000 years is 5 grams, how much radioactive carbon did you start with?"

equation given to me is $A = A_0e^{-kt}$

Now how do I find $A_0$

I've found $k = -0.000121$

Therefore I have formulated this equation

$5 = A_0e^{(-0.000121*1000)}$

My problem is, how do I find $A_0$ from the above formula or have I done something wrong. Everytime I do it, I end up with weird answer. Please help

Just out of curiosity though, my answer = 5.643124562. Is that right? or wrong?

**artvandalay11** · June 7th 2009, 06:17 PM

you know how to solve that, as you did since you got a correct answer

**mr fantastic** · June 7th 2009, 06:22 PM

Originally Posted by dwat

"The half-life of radioactive carbon (14) is 5730 years. If the amount of radioactive carbon left over after 1000 years is 5 grams, how much radioactive carbon did you start with?"

equation given to me is $A = A_0e^{-kt}$

Now how do I find $A_0$

I've found $k = -0.000121$ Mr F says: There should be no negative. Clearly a typo on your part. And you should state the accuracy that you've rounded your answer to.

Therefore I have formulated this equation

$5 = A_0e^{(-0.000121*1000)}$

My problem is, how do I find $A_0$ from the above formula or have I done something wrong. Everytime I do it, I end up with weird answer. Please help

Just out of curiosity though, my answer = 5.643124562. Is that right? or wrong?

$5 = A_0 e^{-(0.000121)(1000)} = A_0 e^{-0.121} \Rightarrow A_0 = \frac{5}{e^{-0.121}} = 5.64$ gram, correct to three significant figures.

Since you have rounded your calculated value of k to three significant figures, your final answer is calculator madness - you cannot have an answer that is more accurate than the least accurate data used to get the answer ....

**dwat** · June 7th 2009, 07:49 PM

Originally Posted by mr fantastic

$5 = A_0 e^{-(0.000121)(1000)} = A_0 e^{-0.121} \Rightarrow A_0 = \frac{5}{e^{-0.121}} = 5.64$ gram, correct to three significant figures.

Since you have rounded your calculated value of k to three significant figures, your final answer is calculator madness - you cannot have an answer that is more accurate than the least accurate data used to get the answer ....

So I did get it right

Oh sorry, yeah I meant k = 0.000121

Well it asked to round k to 6 decimal places so I did. But you have a point. My mistake

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